Two blocks are in contact on a frictionless table masses $2m$ and $m$ respectively as shown in figure. Force F is applied on mass $2m$ then system moves towards right. Now the same force F is applied on $m$. The ratio of force of contact between the two blocks will be in the two cases respectively.

The correct option is (C) 1:2

$Asweknow{F}_{net}=m_{net}\times a_{net}$

$\therefore a_{net}=\frac{F_n}{m_n}$

$case\left(1\right)Asweapplyforceon2m,theentiresystemtendtoacceleratetowardsrightside,$

$resultingnetaccelerationofthesystem,a_{net}=\frac{F}{2m+m}\Rightarrow \frac{F}{3m}$

each block move with the same acceleration as the surface is frictionless.

Hint: Draw FBD for better understanding.

lets choose the body on which less number of forces are acting, as it makes the process quicker.

$formmassofbody,$

$N_1=m\times a_{net}$

$N_1=m\left(\frac{F}{3m}\right)$

$N_1=\frac{F}{3}$

$similarlyforcase\left(2\right)whereforceFactsonmmassandtheentriresystemacceleratetowardsleft.$

$DrawFBD,tryyourselfasapartofchallenge.$

$weget$

$N_2=2m\times a_{net}$

$N_2=2m\left(\frac{F}{3m}\right)$

$N_2=\frac{2F}{3}$

$finally\frac{N_1}{N_2}=\frac{1}{2}$